已知 $\text{e}^y = 2 + \text{e}^x$。

1. Differentiate the given equation implicitly with respect to $x$.

Given:
$\text{e}^y = 2 + \text{e}^x$
Differentiating both sides with respect to $x$:
$\text{e}^y \frac{\text{d}y}{\text{d}x} = \text{e}^x$

2. Differentiate the resulting equation a second time with respect to $x$.

Applying the product rule on the left-hand side:
$\frac{\text{d}}{\text{d}x}\left(\text{e}^y\right) \cdot \frac{\text{d}y}{\text{d}x} + \text{e}^y \cdot \frac{\text{d}}{\text{d}x}\left(\frac{\text{d}y}{\text{d}x}\right) = \frac{\text{d}}{\text{d}x}(\text{e}^x)$
$\text{e}^y \left(\frac{\text{d}y}{\text{d}x}\right)^2 + \text{e}^y \frac{\text{d}^2y}{\text{d}x^2} = \text{e}^x$

3. Substitute into the second derivative equation to show the required differential equation.

From the first derivative, we know $\text{e}^x = \text{e}^y \frac{\text{d}y}{\text{d}x}$. Substituting this into the right-hand side of our second derivative equation gives:
$\text{e}^y \frac{\text{d}^2y}{\text{d}x^2} + \text{e}^y \left(\frac{\text{d}y}{\text{d}x}\right)^2 = \text{e}^y \frac{\text{d}y}{\text{d}x}$
Since $\text{e}^y \neq 0$, we can divide the entire equation by $\text{e}^y$:
$\frac{\text{d}^2y}{\text{d}x^2} + \left(\frac{\text{d}y}{\text{d}x}\right)^2 = \frac{\text{d}y}{\text{d}x} \quad \text{(shown)}$

4. Evaluate $y$, the first derivative, and the second derivative at $x = 0$.

When $x = 0$:
$\text{e}^y = 2 + \text{e}^0 = 2 + 1 = 3 \implies y = \ln 3$
Substitute $x = 0$ and $y = \ln 3$ into the first derivative equation:
$\text{e}^{\ln 3} \frac{\text{d}y}{\text{d}x} = \text{e}^0 \implies 3\frac{\text{d}y}{\text{d}x} = 1 \implies \frac{\text{d}y}{\text{d}x} = \frac{1}{3}$
Substitute $\frac{\text{d}y}{\text{d}x} = \frac{1}{3}$ into the shown differential equation:
$\frac{\text{d}^2y}{\text{d}x^2} + \left(\frac{1}{3}\right)^2 = \frac{1}{3}$
$\frac{\text{d}^2y}{\text{d}x^2} = \frac{1}{3} - \frac{1}{9} = \frac{2}{9}$

5. Construct the Maclaurin series using the evaluated derivatives.

The Maclaurin series formula is:
$y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \dots$
Substituting the exact values found:
$y = \ln 3 + \frac{1}{3}x + \frac{\left(\frac{2}{9}\right)}{2}x^2$
$y = \ln 3 + \frac{1}{3}x + \frac{1}{9}x^2$

1. Express $y$ explicitly and substitute the standard expansion of $\text{e}^x$.

Given $\text{e}^y = 2 + \text{e}^x$, we take the natural logarithm on both sides:
$y = \ln(2 + \text{e}^x)$
Using the standard series $\text{e}^x \approx 1 + x + \frac{x^2}{2}$ up to the $x^2$ term:
$y \approx \ln\left(2 + 1 + x + \frac{x^2}{2}\right) = \ln\left(3 + x + \frac{x^2}{2}\right)$

2. Factor out $3$ to prepare for the $\ln(1+u)$ standard series expansion.

$y \approx \ln\left[3\left(1 + \frac{x}{3} + \frac{x^2}{6}\right)\right]$
Using the logarithm property $\ln(ab) = \ln a + \ln b$:
$y \approx \ln 3 + \ln\left(1 + \frac{x}{3} + \frac{x^2}{6}\right)$

3. Apply the standard expansion for $\ln(1+u)$ and simplify to verify the series.

Using the standard series $\ln(1+u) \approx u - \frac{u^2}{2}$ with $u = \frac{x}{3} + \frac{x^2}{6}$:
$y \approx \ln 3 + \left(\frac{x}{3} + \frac{x^2}{6}\right) - \frac{1}{2}\left(\frac{x}{3} + \frac{x^2}{6}\right)^2$
Discarding terms of degree higher than $2$:
$y \approx \ln 3 + \frac{x}{3} + \frac{x^2}{6} - \frac{1}{2}\left(\frac{x^2}{9}\right)$
$y \approx \ln 3 + \frac{1}{3}x + \left(\frac{1}{6} - \frac{1}{18}\right)x^2$
$y \approx \ln 3 + \frac{1}{3}x + \left(\frac{3}{18} - \frac{1}{18}\right)x^2$
$y \approx \ln 3 + \frac{1}{3}x + \frac{1}{9}x^2$
This identical result verifies the correctness of the series expansion found in part (a).